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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Differentiate w.r.t. x the function in Exercise:
$(\log\text{x})^{\log\text{x}},\text{x}>1$
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Answer

Let $\text{y}=(\log\text{x})^{\log\text{x}}$
Tanking logarithm on both the sides, we obtain
$\log\text{y}=\log\text{x}.\log(\log\text{x})$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}[\log\text{x}.\log(\log\text{x})]$
$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(\log\text{x)}.\frac{\text{d}}{\text{dx}}(\log\text{x)}+\log\text{x}.\frac{\text{d}}{\text{dx}}[\log(\log\text{x})]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(\log\text{x)}.\frac{1}{\text{x}}+\log\text{x}.\frac{1}{\log\text{x}}.\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\text{x}}\log(\log\text{x)}+\frac{1}{\text{x}}\Big]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=(\log\text{x)}^{\log\text{x}}\Big[\frac{1}{\text{x}}+\frac{\log(\log\text{x})}{\text{x}}\Big]$

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